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# bijection between closed and open interval

For mapping the end points of the closed unit interval $[0,\,1]$ and its inner points bijectively onto the corresponding open unit interval $(0,\,1)$, one has to discern suitable denumerable subsets in both sets:

$\displaystyle[0,\,1]\,\;=\;\{0,\,1,\,1/2,\,1/3,\,1/4,\,\ldots\}\cup S,$ | |||

$\displaystyle(0,\,1)\;=\;\{1/2,\,1/3,\,1/4,\,\ldots\}\cup S,$ |

where

$S\;:=\;[0,\,1]\smallsetminus\{0,\,1,\,1/2,\,1/3,\,1/4,\,\ldots\}.$ |

Then the mapping $f$ from $[0,\,1]$ to $(0,\,1)$ defined by

$f(x)\;:=\;\begin{cases}1/2\quad\mbox{for}\quad x=0,\\ 1/(n\!+\!2)\quad\mbox{for}\quad x=1/n\quad(n=1,\,2,\,3,\,\ldots),\\ x\qquad\mbox{for}\quad x\in S\end{cases}$ |

is apparently a bijection. This means the equicardinality of both intervals.

Note that the bijection is neither monotonic (e.g. $0\mapsto\frac{1}{2}$, $\frac{1}{2}\mapsto\frac{1}{4}$, $1\mapsto\frac{1}{3}$) nor continuous. Generally, there does not exist any continuous surjective mapping $[0,\,1]\to(0,\,1)$, since by the intermediate value theorem a continuous function maps a closed interval to a closed interval.

# References

- 1 S. Lipschutz: Set theory. Schaum Publishing Co., New York (1964).

Related:

BijectionBetweenUnitIntervalAndUnitSquare

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## Mathematics Subject Classification

54C30*no label found*26A30

*no label found*

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